-5t^2+10t+18=18t

Solution for -5t^2+10t+18=18t equation:

-5t^2+10t+18=18t

We move all terms to the left:

-5t^2+10t+18-(18t)=0

We add all the numbers together, and all the variables

-5t^2-8t+18=0

a = -5; b = -8; c = +18;
Δ = b2-4ac
Δ = -82-4·(-5)·18
Δ = 424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{106}}{2*-5}=\frac{8-2\sqrt{106}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{106}}{2*-5}=\frac{8+2\sqrt{106}}{-10}
$